I'm currently going through Spivak and ran across this problem, but i see a difference in my answer and the answer that i'm checking it again.
The problem is to eliminate the absolute value signs in $|(|x|-1)|$
In my approach, i take two cases: $|x| < 1$ or $|x| \ge 1$
Evaluating the first case we get $-(x-1)$ and evaluating the second we get $x-1$.
In the answer that i'm comparing it against however,
|(|x|-1)| This becomes |x|-1 and 1-|x|.
|x|-1 can be either x-1 or -x-1.
1-|x| can be either 1-x or x+1.
Four possibilities.
I don't see why we should evaluate the outer absolute block first
hint
$$||x|-1|= \begin{cases} |x|-1 & \text{if } |x| \geq 1\\ -(|x|-1) & \text{if } |x| < 1. \end{cases} $$ Now do two cases, namely $x \geq 0$ and $x <0$ for each part of the piecewise function.